6
$\begingroup$

Let $\Phi$ be a Youngs's function, i.e. $$ \Phi(t) = \int_0^t \varphi(s) \,\mathrm d s$$ for some $\varphi$ satifying

  1. $\varphi:[0,\infty)\to[0,\infty]$ is increasing
  2. $\varphi$ is lower semi continuous
  3. $\varphi(0) = 0$
  4. $\varphi$ is neither identically zero nor identically infinite

and define the Luxemburg norm of $f:\Omega\to\mathbb{R}$ as $$ \lVert f \rVert_{L^{\Phi}} := \inf \left\{\gamma\,\middle|\,\gamma>0,\,\int_{\Omega} \Phi\left(\frac {\lvert f(x)\rvert}{\gamma} \right)\,\mathrm{d}x\leq 1\right\}.$$


Question: What can we say about $\Phi\left(\lVert f \rVert_{L^{\Phi}}\right)$? In particular, I'd like to know, if $$\Phi\left(\lVert f \rVert_{L^{\Phi}}\right) \leq C \int_{\Omega}\Phi(\lvert f(x)\rvert) \,\mathrm d x$$ holds for some $C$ independent of $f$.

Any idea or hint for a reference is welcome!


Notes:

  • The above inequality trivially holds for $\Phi(t) = t^p$, where $p>1$
  • Maybe it's appropriate to consider this question in the more general framework of Musielak-Orlicz spaces. However, e.g. in Lebesgue and Sobolev Spaces with Variable Exponents I was unable to find an appropriate result.
  • Since there has been no response yet, I also asked the question on MathOverflow.
$\endgroup$
3
  • $\begingroup$ Do you mean to write $\inf \{\int_\Omega\dots | \gamma>0\}$ instead of $\inf \{\gamma>0 | \int_\Omega\dots \}$? The notation is not clear to me $\endgroup$
    – harfe
    Jun 13, 2019 at 9:02
  • $\begingroup$ No, it's supposed to be the $\inf$ over the $\gamma$. Maybe $\inf\{\gamma\,|\,\gamma>0,\,\int_{\Omega}...\}$ is better? Ah, and I forgot the $\leq 1$ … Sorry! $\endgroup$
    – CallMeStag
    Jun 13, 2019 at 9:26
  • $\begingroup$ A counterexample has been given by @harfe on MathOverflow. $\endgroup$
    – CallMeStag
    Jun 17, 2019 at 6:25

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.